A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be _________ rad s⁻².

Solution

Physics sir · 3 steps

Rope tension gives the torque, the hollow cylinder's MR² resists it, and α is just their ratio.

1. Torque from rope. The rope leaves the cylinder along a tangent, so the force acts at a perpendicular distance R = 0.7 m from the axis. Torque τ = FR = 52.5 × 0.7 = 36.75 N m.
2. Moment of inertia. In a hollow cylinder all the mass sits at distance R from the axis, so I = MR² = 5 × (0.7)² = 2.45 kg m².

   α = τ/I = FR/MR² = F/MR = 52.5/(5 × 0.7)

3. So. α = 52.5/3.5 = 15 rad s⁻², so the answer is 15.

The common slip: 30. Using the solid-cylinder value I = 1/2MR² doubles the answer to 30. A hollow cylinder has I = MR².

The idea to keep. For a rope pulled off any rim, τ = FR. Then pick the right I: MR² for a hollow cylinder or ring, 1/2MR² for a solid one. Always convert cm to m first.

Answer from the NTA final answer key. Solution by PrepWiser.